A stock car starts from rest at time t=0 with velocity (m/s) increasing for 4.0 s , according to the function Vx=1.4t2+1.1t.
a) Find the car's velocity at the end of the 4.0 s interval.
b) Find the average acceleration for this interval.
a)assume you mean Vx=1.4t^2+1.1t
v = 1.4(16) + 1.1(4)
b) change in velocity
= answer to part a since velocity at t = 0 was 0
so
divide a the answer to part a by 4 seconds
To find the velocity of the car at the end of the 4.0 s interval, we can use the function Vx=1.4t^2+1.1t, where t is the time elapsed. We need to find the value of Vx when t is equal to 4.0 s.
a) To find the velocity at the end of the 4.0 s interval, substitute t=4.0 s into the function Vx=1.4t^2+1.1t:
Vx = 1.4(4.0)^2 + 1.1(4.0)
= 1.4(16.0) + 4.4
= 22.4 + 4.4
= 26.8 m/s
Therefore, the car's velocity at the end of the 4.0 s interval is 26.8 m/s.
b) To find the average acceleration for this interval, we need to calculate the change in velocity (ΔVx) over the given time interval (Δt).
The change in velocity (ΔVx) can be found by subtracting the initial velocity (Vx_i) from the final velocity (Vx_f). In this case, the car starts from rest, so the initial velocity is 0 m/s.
ΔVx = Vx_f - Vx_i
= 26.8 m/s - 0 m/s
= 26.8 m/s
The change in time (Δt) is given as 4.0 s.
Now, we can calculate the average acceleration (a_avg) using the formula:
a_avg = ΔVx / Δt
a_avg = 26.8 m/s / 4.0 s
= 6.7 m/s^2
Therefore, the average acceleration for this interval is 6.7 m/s^2.